我在这样的文件中有一个版本号:
测试x.x.x.x
所以我像这样抓它:
import re
def increment(match):
# convert the four matches to integers
a,b,c,d = [int(x) for x in match.groups()]
# return the replacement string
return f'{a}.{b}.{c}.{d}'
lines = open('file.txt', 'r').readlines()
lines[3] = re.sub(r"\b(\d+)\.(\d+)\.(\d+)\.(\d+)\b", increment, lines[3])
我要这样做,如果最后一位是9
...,然后将其更改为0
,然后将前一位更改为1。所以1.1.1.9
更改为1.1.2.0
。
我是这样做的:
def increment(match):
# convert the four matches to integers
a,b,c,d = [int(x) for x in match.groups()]
# return the replacement string
if (d == 9):
return f'{a}.{b}.{c+1}.{0}'
elif (c == 9):
return f'{a}.{b+1}.{0}.{0}'
elif (b == 9):
return f'{a+1}.{0}.{0}.{0}'
问题在其1.1.9.9
或1.9.9.9
时发生。需要四舍五入的地方。我该如何处理这个问题?
答案 0 :(得分:3)
使用整数加法吗?
def increment(match):
# convert the four matches to integers
a,b,c,d = [int(x) for x in match.groups()]
*a,b,c,d = [int(x) for x in str(a*1000 + b*100 + c*10 + d + 1)]
a = ''.join(map(str,a)) # fix for 2 digit 'a'
# return the replacement string
return f'{a}.{b}.{c}.{d}'
答案 1 :(得分:1)
如果您的版本永远不会超过10,则最好将其转换为整数,将其递增,然后再转换回字符串。 这使您可以根据需要增加任意数量的版本号,而不仅限于数千个。
def increment(match):
match = match.replace('.', '')
match = int(match)
match += 1
match = str(match)
output = '.'.join(match)
return output
答案 2 :(得分:0)
将1
添加到最后一个元素。如果大于9
,请将其设置为0
,并对上一个元素执行相同的操作。视需要重复:
import re
def increment(match):
# convert the four matches to integers
g = [int(x) for x in match.groups()]
# increment, last one first
pos = len(g)-1
g[pos] += 1
while pos > 0:
if g[pos] > 9:
g[pos] = 0
pos -= 1
g[pos] += 1
else:
break
# return the replacement string
return '.'.join(str(x) for x in g)
print (re.sub(r"\b(\d+)\.(\d+)\.(\d+)\.(\d+)\b", increment, '1.8.9.9'))
print (re.sub(r"\b(\d+)\.(\d+)\.(\d+)\.(\d+)\b", increment, '1.9.9.9'))
print (re.sub(r"\b(\d+)\.(\d+)\.(\d+)\.(\d+)\b", increment, '9.9.9.9'))
结果:
1.9.0.0
2.0.0.0
10.0.0.0